将一个圆分成 n 块作为大小为 n 的数组。数组的第 i 个元素表示一件的角度。我们的任务是从这些部分制作两个连续的部分，使这两个部分的角度差异最小。
例子 ：
Input : arr[] = {90, 90, 90, 90}
Output : 0
In this example, we can take 1 and 2
pieces and 3 and 4 pieces. Then the
answer is |(90 + 90) - (90 + 90)| = 0.
Input : arr[] = {170, 30, 150, 10}
Output : 0
In this example, we can take 1 and 4,
and 2 and 3 pieces. So the answer is
|(170 + 10) - (30 + 150)| = 0.
Input : arr[] = {100, 100, 160}
Output : 40我们可以注意到，如果其中一个部分是连续的，那么所有剩余的部分也形成一个连续的部分。如果第一部分的角度等于 x，则第一部分和第二部分的角度差为 |x – (360 – x)| = |2 * x – 360| = 2 * |x – 180|。所以对于每个可能的连续部分，我们可以计算它的角度并更新答案。
C++
// CPP program to find minimum
// difference of angles of two
// parts of given circle.
#include
using namespace std;
// Returns the minimum difference
// of angles.
int findMinimumAngle(int arr[], int n)
{
int l = 0, sum = 0, ans = 360;
for (int i = 0; i < n; i++) {
// sum of array
sum += arr[i];
while (sum >= 180) {
// calculating the difference of
// angles and take minimum of
// previous and newly calculated
ans = min(ans, 2 * abs(180 - sum));
sum -= arr[l];
l++;
}
ans = min(ans, 2 * abs(180 - sum));
}
return ans;
}
// driver code
int main()
{
int arr[] = { 100, 100, 160 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findMinimumAngle(arr, n) << endl;
return 0;
}
// This code is contributed by "Abhishek Sharma 44" Java
// java program to find minimum
// difference of angles of two
// parts of given circle.
import java.util.*;
class Count{
public static int findMinimumAngle(int arr[], int n)
{
int l = 0, sum = 0, ans = 360;
for (int i = 0; i < n; i++)
{
// sum of array
sum += arr[i];
while (sum >= 180)
{
// calculating the difference of
// angles and take minimum of
// previous and newly calculated
ans = Math.min(ans,
2 * Math.abs(180 - sum));
sum -= arr[l];
l++;
}
ans = Math.min(ans,
2 * Math.abs(180 - sum));
}
return ans;
}
public static void main(String[] args)
{
int arr[] = { 100, 100, 160 };
int n = 3;
System.out.print(findMinimumAngle(arr, n));
}
}
// This code is contributed by rishabh_jainPython3
# java program to find minimum
# difference of angles of two
# parts of given circle.
import math
# function returns the minimum
# difference of angles.
def findMinimumAngle (arr, n):
l = 0
_sum = 0
ans = 360
for i in range(n):
#sum of array
_sum += arr[i]
while _sum >= 180:
# calculating the difference of
# angles and take minimum of
# previous and newly calculated
ans = min(ans, 2 * abs(180 - _sum))
_sum -= arr[l]
l+=1
ans = min(ans, 2 * abs(180 - _sum))
return ans
# driver code
arr = [100, 100, 160]
n = len(arr)
print(findMinimumAngle (arr, n))
# This code is contributed by "Abhishek Sharma 44"C#
// C# program to find minimum
// difference of angles of two
// parts of given circle.
using System;
class GFG
{
public static int findMinimumAngle(int []arr, int n)
{
int l = 0, sum = 0, ans = 360;
for (int i = 0; i < n; i++)
{
// sum of array
sum += arr[i];
while (sum >= 180)
{
// calculating the difference of
// angles and take minimum of
// previous and newly calculated
ans = Math.Min(ans,
2 * Math.Abs(180 - sum));
sum -= arr[l];
l++;
}
ans = Math.Min(ans,
2 * Math.Abs(180 - sum));
}
return ans;
}
// Driver code
public static void Main()
{
int []arr = { 100, 100, 160 };
int n = 3;
Console.WriteLine(findMinimumAngle(arr, n));
}
}
// This code is contributed by vt_mPHP
= 180)
{
// calculating the difference of
// angles and take minimum of
// previous and newly calculated
$ans = min($ans, 2 *
abs(180 - $sum));
$sum -= $arr[$l];
$l++;
}
$ans = min($ans, 2 * abs(180 - $sum));
}
return $ans;
}
// Driver Code
$arr = array( 100, 100, 160 );
$n = sizeof($arr);
echo findMinimumAngle($arr, $n), "\n" ;
// This code is contributed by m_kit
?>Javascript
输出 ：
40